Key Concepts and Formulas

• Area between curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $A = \int_a^b [f(x) - g(x)] \, dx$.
• Inverse Trigonometric Integrals: Recall that $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$.
• Symmetry: Utilize symmetry about the x-axis to simplify area calculations.

Step-by-Step Solution

Step 1: Find the points of intersection

We need to find where the parabola $y^2 = 4x$ and the circle $x^2 + y^2 = 5$ intersect. Substituting $y^2 = 4x$ into the equation of the circle, we get: $x^2 + 4x = 5$ $x^2 + 4x - 5 = 0$ $(x+5)(x-1) = 0$ So, $x = -5$ or $x = 1$. Since $y^2 = 4x$, $x$ cannot be negative. Thus, $x = 1$. When $x = 1$, $y^2 = 4(1) = 4$, so $y = \pm 2$. Therefore, the points of intersection are $(1, 2)$ and $(1, -2)$.

Step 2: Visualize the region and set up the integral

The circle $x^2 + y^2 = 5$ has radius $\sqrt{5}$ and is centered at the origin. The parabola $y^2 = 4x$ opens to the right with its vertex at the origin. We want to find the area of the smaller region enclosed by the circle and the parabola. This region is symmetric about the x-axis. We will calculate the area of the upper half and then double it.

The equation of the circle can be written as $x = \sqrt{5 - y^2}$ (for the right half, which bounds the region). The equation of the parabola can be written as $x = \frac{y^2}{4}$. We will integrate with respect to $y$.

Step 3: Calculate the area of the upper half of the region

The upper half of the smaller area is bounded by $x = \sqrt{5 - y^2}$ and $x = \frac{y^2}{4}$ from $y = 0$ to $y = 2$. The area of this region is:

$A_{half} = \int_0^2 \left(\sqrt{5 - y^2} - \frac{y^2}{4}\right) \, dy$

We split the integral: $A_{half} = \int_0^2 \sqrt{5 - y^2} \, dy - \int_0^2 \frac{y^2}{4} \, dy$

Step 4: Evaluate the integrals

Let's evaluate the first integral: $\int_0^2 \sqrt{5 - y^2} \, dy$ Using the formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$ with $a = \sqrt{5}$, we get:

$\int_0^2 \sqrt{5 - y^2} \, dy = \left[\frac{y}{2}\sqrt{5 - y^2} + \frac{5}{2} \sin^{-1}\left(\frac{y}{\sqrt{5}}\right)\right]_0^2$ $= \left(\frac{2}{2}\sqrt{5 - 4} + \frac{5}{2} \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)\right) - \left(0 + \frac{5}{2} \sin^{-1}(0)\right)$ $= 1 + \frac{5}{2} \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$

Now, let's evaluate the second integral: $\int_0^2 \frac{y^2}{4} \, dy = \left[\frac{y^3}{12}\right]_0^2 = \frac{2^3}{12} - \frac{0^3}{12} = \frac{8}{12} = \frac{2}{3}$

So, $A_{half} = 1 + \frac{5}{2} \sin^{-1}\left(\frac{2}{\sqrt{5}}\right) - \frac{2}{3} = \frac{1}{3} + \frac{5}{2} \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$

Step 5: Calculate the total area

Since the region is symmetric about the x-axis, the total area is twice the area of the upper half: $A_{total} = 2 \cdot A_{half} = 2 \left(\frac{1}{3} + \frac{5}{2} \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$ $A_{total} = \frac{2}{3} + 5 \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$

Common Mistakes & Tips

• Sign Errors: Be careful with signs when substituting and evaluating limits of integration.
• Incorrect Formula: Make sure you are using the correct integral formula for $\int \sqrt{a^2 - x^2} \, dx$.
• Symmetry: Utilize symmetry to simplify the calculations by only finding the area of half the region and doubling it.

Summary

To find the area of the smaller region enclosed by the parabola $y^2 = 4x$ and the circle $x^2 + y^2 = 5$, we first found the intersection points of the two curves. We then set up a definite integral to represent the area of the upper half of the region, taking advantage of the symmetry about the x-axis. We evaluated the integral using standard integration techniques, including the formula for integrating $\sqrt{a^2 - x^2}$. Finally, we doubled the area of the upper half to find the total area of the smaller region, which is $\frac{2}{3} + 5 \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$.

Final Answer The final answer is $\boxed{\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)}$, which corresponds to option (A).